Linear Programming: Sensitivity Analysis

and Interpretation of Solution Pertemuan 5

Matakuliah : K0442-Metode KuantitatifTahun : 2009

Material Outline

• Introduction to Sensitivity Analysis• Graphical Sensitivity Analysis• Sensitivity Analysis: Computer Solution• Simultaneous Changes

Sensitivity Analysis• Sensitivity analysis (or post-optimality analysis) is

used to determine how the optimal solution is affected by changes, within specified ranges, in:– the objective function coefficients– the right-hand side (RHS) values

• Sensitivity analysis is important to the manager who must operate in a dynamic environment with imprecise estimates of the coefficients.

• Sensitivity analysis allows him to ask certain what-if questions about the problem.

Example 1• LP Formulation

Max z = 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1, x2 > 0

Example 1

• Graphical Solution

88

77

66

55

44

33

22

11

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10

2x1 + 3x2 < 19

x2

xx11

x1 + x2 < 8Max 5x1 + 7x2Max 5x1 + 7x2

x1 < 6

Optimal: x1 = 5, x2 = 3, z = 46

Objective Function Coefficients• Let us consider how changes in the objective

function coefficients might affect the optimal solution.

• The range of optimality for each coefficient provides the range of values over which the current solution will remain optimal.

• Managers should focus on those objective coefficients that have a narrow range of optimality and coefficients near the endpoints of the range.

Example 1• Changing Slope of Objective Function

88

77

66

55

44

33

22

11

1 2 3 4 5 6 7 8 9 101 2 3 4 5 6 7 8 9 10 xx11

FeasibleFeasibleRegionRegion

1111 2222

3333

4444

5555

Range of Optimality• Graphically, the limits of a range of optimality are

found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines.

• The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2, and the slope of a constraint, a1x1 + a2x2 = b, is -a1/a2.

Example 1• Range of Optimality for c1

The slope of the objective function line is -c1/c2. The slope of the first binding constraint, x1 + x2 = 8, is -1 and the slope of the second binding constraint, x1 + 3x2 = 19, is -2/3.

Find the range of values for c1 (with c2 staying 7) such that the objective function line slope lies between that of the two binding constraints:

-1 < -c1/7 < -2/3

Multiplying through by -7 (and reversing the inequalities): 14/3 < c1 < 7

Example 1• Range of Optimality for c2

Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints:

-1 < -5/c2 < -2/3

Multiplying by -1: 1 > 5/c2 > 2/3

Inverting, 1 < c2/5 < 3/2

Multiplying by 5: 5 < c2 < 15/2

Right-Hand Sides• Let us consider how a change in the right-hand side for a

constraint might affect the feasible region and perhaps cause a change in the optimal solution.

• The improvement in the value of the optimal solution per unit increase in the right-hand side is called the dual price.

• The range of feasibility is the range over which the dual price is applicable.

• As the RHS increases, other constraints will become binding and limit the change in the value of the objective function.

Dual Price• Graphically, a dual price is determined by adding +1 to

the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints.

• The dual price is equal to the difference in the values of the objective functions between the new and original problems.

• The dual price for a nonbinding constraint is 0.• A negative dual price indicates that the objective

function will not improve if the RHS is increased.

Relevant Cost and Sunk Cost

• A resource cost is a relevant cost if the amount paid for it is dependent upon the amount of the resource used by the decision variables.

• Relevant costs are reflected in the objective function coefficients.

• A resource cost is a sunk cost if it must be paid regardless of the amount of the resource actually used by the decision variables.

• Sunk resource costs are not reflected in the objective function coefficients.

A Cautionary Noteon the Interpretation of Dual Prices

• Resource cost is sunkThe dual price is the maximum amount you should be willing to pay for one additional unit of the resource.

• Resource cost is relevantThe dual price is the maximum premium over the normal cost that you should be willing to pay for one unit of the resource.

Example 1• Dual Prices Constraint 1: Since x1 < 6 is not a binding

constraint, its dual price is 0. Constraint 2: Change the RHS value of the second

constraint to 20 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 20 and x1 + x2 = 8.

The solution is x1 = 4, x2 = 4, z = 48. Hence, the dual price = znew - zold = 48 - 46 = 2.

Example 1• Dual Prices

Constraint 3: Change the RHS value of the third constraint to 9 and resolve for the

optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9.

The solution is: x1 = 8, x2 = 1, z = 47. Hence, the dual price is znew - zold = 47 - 46 = 1.

Range of Feasibility• The range of feasibility for a change in the right

hand side value is the range of values for this coefficient in which the original dual price remains constant.

• Graphically, the range of feasibility is determined by finding the values of a right hand side coefficient such that the same two lines that determined the original optimal solution continue to determine the optimal solution for the problem.

Range of Optimality and 100% Rule• The 100% rule states that simultaneous changes

in objective function coefficients will not change the optimal solution as long as the sum of the percentages of the change divided by the corresponding maximum allowable change in the range of optimality for each coefficient does not exceed 100%.

Range of Feasibility and 100% Rule• The 100% rule states that simultaneous changes

in right-hand sides will not change the dual prices as long as the sum of the percentages of the changes divided by the corresponding maximum allowable change in the range of feasibility for each right-hand side does not exceed 100%.

Example 2

• Consider the following linear program:

Min 6x1 + 9x2 ($ cost)

s.t. x1 + 2x2 < 8

10x1 + 7.5x2 > 30

x2 > 2

x1, x2 > 0

Example 2

• The Management Scientist Output

OBJECTIVE FUNCTION VALUE = 27.000

Variable Value Reduced Cost x1 1.500 0.000

x2 2.000 0.000

Constraint Slack/Surplus Dual Price 1 2.500 0.000 2 0.000 -0.600

3 0.000 -4.500

Example 2• The Management Scientist Output (Continued)

OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit x1 0.000 6.000 12.000

x2 4.500 9.000 No Upper Limit

RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 5.500 8.000 No Upper Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

Example 2• Optimal Solution

According to the output: x1 = 1.5

x2 = 2.0

Objective function value = 27.00

Example 2• Range of Optimality

QuestionSuppose the unit cost of x1 is decreased to

$4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4?

Example 2OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit

x1 0.000 6.000 12.000

x2 4.500 9.000 No Upper Limit

RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit

1 5.500 8.000 No Upper Limit

2 15.000 30.000 55.000 3 0.000 2.000 4.000

Example 2• Range of Optimality

AnswerThe output states that the solution remains

optimal as long as the objective function coefficient of x1 is between 0 and 12. Since 4 is within this range, the optimal solution will not change. However, the optimal total cost will be affected: 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.

Example 2

• Range of OptimalityQuestion

How much can the unit cost of x2 be decreased without concern for the optimal solution changing?

Example 2

OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit

x1 0.000 6.000 12.000

x2 4.500 9.000 No Upper Limit

RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit

1 5.500 8.000 No Upper Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

Example 2• Range of Optimality

AnswerThe output states that the solution remains

optimal as long as the objective function coefficient of x2 does not fall below 4.5.

Example 2• Range of Optimality and 100% Rule

QuestionIf simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was reduced to $6, would the current solution remain optimal?AnswerIf c1 = 7.5, the amount c1 changed is 7.5 - 6 = 1.5. The maximum allowable increase is 12 - 6 = 6, so this is a 1.5/6 = 25% change. If c2 = 6, the amount that c2 changed is 9 - 6 = 3. The maximum allowable decrease is 9 - 4.5 = 4.5, so this is a 3/4.5 = 66.7% change. The sum of the change percentages is 25% + 66.7% = 91.7%. Since this does not exceed 100% the optimal solution would not change.

Example • Range of Feasibility

QuestionIf the right-hand side of constraint 3 is

increased by 1, what will be the effect on the optimal solution?

Example 2OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit x1 0.000 6.000 12.000

x2 4.500 9.000 No Upper Limit

RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit

1 5.500 8.000 No Upper Limit 2 15.000 30.000 55.000 3 0.000 2.000 4.000

Example 2• Range of Feasibility

AnswerA dual price represents the improvement in the

objective function value per unit increase in the right-hand side. A negative dual price indicates a deterioration (negative improvement) in the objective, which in this problem means an increase in total cost because we're minimizing. Since the right-hand side remains within the range of feasibility, there is no change in the optimal solution. However, the objective function value increases by $4.50.